Chapter 16, Problem 16.12P

(a)

Interpretation Introduction

Interpretation:

The colligative molality of a solution prepared by dissolving one mole of methanol in $1.0kg$ of water has to be calculated.

Concept Introduction:

Molality:

Molality is defined as the number of moles of solute per $1000g$ or one kilogram of solvent.  Mathematically, it can be represented as given below.

  $Molality=\frac{Molesofsolute}{1kgofsolvent}$

The colligative molality of a solution is the product of number of solute particles per formula unit and molality of solution.  This can be represented as given below.

  $m_c=i\times m$

Where, $i$ is the number of solute particles produced per formula unit when the solute is dissolved in the solvent.

Answer to Problem 16.12P

The colligative molality of the resulting solution is $1.00m_c.$

Explanation of Solution

Given that, one mole of methanol is dissolved $1.0kg$ of water.  Here, solute is methanol and solvent is water.

The molality of the resulting solution can be calculated as shown below.

  $\begin{array}{l}Molality=\frac{Molesofsolute}{1kgofsolvent}\\ \\ Molality=\frac{1mol}{1.0kg}=1.00m.\end{array}$

For methanol, $i$ is one.

Then, the colligative molality will be same as the molality.

  $m_c=i\times m=1\times 1.00m=1.00m_c.$

Therefore, the colligative molality of the resulting solution is $1.00m_c.$

(b)

Interpretation Introduction

Interpretation:

The colligative molality of a solution prepared by dissolving one mole of $Al{\left(N{O}_3\right)}_3\left(s\right)$ in $1.0kg$ of water has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 16.12P

The colligative molality of the resulting solution is $4.00m_c.$

Explanation of Solution

Given that, one mole of $Al{\left(N{O}_3\right)}_3\left(s\right)$ is dissolved in $1.0kg$ of water.  Here, solute is $Al{\left(N{O}_3\right)}_3\left(s\right)$ and solvent is water.

The molality of the resulting solution can be calculated as shown below.

  $\begin{array}{l}Molality=\frac{Molesofsolute}{1kgofsolvent}\\ \\ Molality=\frac{1mol}{1.0kg}=1.00m.\end{array}$

$Al{\left(N{O}_3\right)}_3\left(s\right)$ is a strong electrolyte and it produces one mole of $A{g}^{+}\left(aq\right)$ ion and three moles of $N{O}_3{}^{-}\left(aq\right)$ ion.  So $i$ is four.

Then,

  $m_c=i\times m=4\times 1.00m=4.00m_c.$

Therefore, the colligative molality of the resulting solution is $4.00m_c.$

(c)

Interpretation Introduction

Interpretation:

The colligative molality of a solution prepared by dissolving one mole of $Fe{\left(N{O}_3\right)}_2\left(s\right)$ in $1.0kg$ of water has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 16.12P

The colligative molality of the resulting solution is $3.00m_c.$

Explanation of Solution

Given that, one mole of $Fe{\left(N{O}_3\right)}_2\left(s\right)$ is dissolved in $1.0kg$ of water.  Here, solute is $Fe{\left(N{O}_3\right)}_2\left(s\right)$ and solvent is water.

The molality of the resulting solution can be calculated as shown below.

  $\begin{array}{l}Molality=\frac{Molesofsolute}{1kgofsolvent}\\ \\ Molality=\frac{1mol}{1.0kg}=1.00m.\end{array}$

$Fe{\left(N{O}_3\right)}_2\left(s\right)$ is a strong electrolyte and it produces one mole of $F{e}^{2+}\left(aq\right)$ ion and two moles of $N{O}_3{}^{-}\left(aq\right)$ ion.  So $i$ is three.

Then,

  $m_c=i\times m=3\times 1.00m=3.00m_c.$

Therefore, the colligative molality of the resulting solution is $3.00m_c.$

(d)

Interpretation Introduction

Interpretation:

The colligative molality of a solution prepared by dissolving one mole of ${\left(N{H}_4\right)}_{2}C{r}_2{O}_7\left(s\right)$ in $1.0kg$ of water has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 16.12P

The colligative molality of the resulting solution is $3.00m_c.$

Explanation of Solution

Given that, one mole of ${\left(N{H}_4\right)}_{2}C{r}_2{O}_7\left(s\right)$ is dissolved in $1.0kg$ of water.  Here, solute is ${\left(N{H}_4\right)}_{2}C{r}_2{O}_7\left(s\right)$ and solvent is water.

The molality of the resulting solution can be calculated as shown below.

  $\begin{array}{l}Molality=\frac{Molesofsolute}{1kgofsolvent}\\ \\ Molality=\frac{1mol}{1.0kg}=1.00m.\end{array}$

${\left(N{H}_4\right)}_{2}C{r}_2{O}_7\left(s\right)$ is a strong electrolyte and it produces two moles of $N{H}_4{}^{+}\left(aq\right)$ ion and one mole of $C{r}_2{O}_7{}^{2-}\left(aq\right)$ ion.  So $i$ is three.

Then,

  $m_c=i\times m=3\times 1.00m=3.00m_c.$

Therefore, the colligative molality of the resulting solution is $3.00m_c.$